This is already fixed in master:
In [1]: srepr(Symbol('x', real=True))
Out[1]: "Symbol('x', real=True)"
Aaron Meurer
On Tue, Jul 21, 2015 at 1:53 PM, G.B. <[email protected]> wrote:
> It looks like srepr() doesn’t capture the real=True parameter to Symbol.
>
> from sympy import *
> s=Symbol('s')
> f=Symbol('f',real=True)
> SExp=Eq(s,2*pi*I*f)
> srepr(SExp)
>
> —> "Equality(Symbol('s'), Mul(Integer(2), I, pi, Symbol('f')))"
>
>
> Is this another issue to be submitted?
>
> In the mean time, I think it’s just going to be most straight forward if I
> convert the iPython notebook example to .py, and attach it. Will that work
> for these other issues?
>
> Thanks—
> Greg
>
>
> On Jul 16, 2015, at 13:32 , Aaron Meurer <[email protected]> wrote:
>
> Oh I didn't realize you had LaTeX strings in your Symbol names. It would
> be more useful to use srepr() then, so that it can just be copy-pasted.
>
> Aaron Meurer
>
> On Thu, Jul 16, 2015 at 12:33 PM, G.B. <[email protected]> wrote:
>
>> Will do. If I print the expression using str(), I get:
>>
>>
>> 'I_{R_2}(s) == -C_{T}*s*(-R_{1}*V_{in}(s)/(R_{1} + 1/(1/(1/(C_{T}*s +
>> 1/(L_{R}*s + R_{2} + 1/(C_{R}*s))) + 1/(C_{L}*s)) + 1/(L_{L}*s))) +
>> V_{in}(s) - (-R_{1}*V_{in}(s)/(R_{1} + 1/(1/(1/(C_{T}*s + 1/(L_{R}*s +
>> R_{2} + 1/(C_{R}*s))) + 1/(C_{L}*s)) + 1/(L_{L}*s))) +
>> V_{in}(s))/(C_{L}*s*(1/(C_{T}*s + 1/(C_{R} + L_{R} + R_{2})) +
>> 1/(C_{L}*s)))) - (-R_{1}*V_{in}(s)/(R_{1} + 1/(1/(1/(C_{T}*s + 1/(L_{R}*s +
>> R_{2} + 1/(C_{R}*s))) + 1/(C_{L}*s)) + 1/(L_{L}*s))) +
>> V_{in}(s))/(1/(C_{T}*s + 1/(C_{R} + L_{R} + R_{2})) + 1/(C_{L}*s))’
>>
>> Is that usable or do I need to provide another means? Is that IPython
>> doc preferable (or the same in straight Python form)?
>>
>> Thanks—
>> Greg
>>
>>
>>
>> On Jul 14, 2015, at 18:05 , Aaron Meurer <[email protected]> wrote:
>>
>> I think there is a bug. If you take the two expressions, subtract them,
>> and call simplify(), you get a result that isn't 0. equals() also says they
>> are different.
>>
>> simplify() returning False is also a bug.
>>
>> Can you open two issues for these things? Just put some minimal code to
>> create the expressions in the issue (i.e., just print the expressions using
>> str()).
>>
>> Aaron Meurer
>>
>> On Tue, Jul 14, 2015 at 12:51 PM, G.B. <[email protected]> wrote:
>>
>>> Hey Aaron—
>>>
>>> Maybe the best way to capture an example is an IPython notebook?
>>>
>>> Let me know if this doesn’t work properly…
>>>
>>> Cheers—
>>>
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>>> > On Jul 13, 2015, at 10:25 , Aaron Meurer <[email protected]> wrote:
>>> >
>>> > On Sun, Jul 12, 2015 at 7:46 PM, G B <[email protected]> wrote:
>>> >> What would case Eq.simplify() to return False? There isn't a lot of
>>> >> computation time before the result.
>>> >
>>> > Can you show an example of what is doing this?
>>> >
>>> >>
>>> >> Also why would using together() before substituting an expression
>>> lead to a
>>> >> numerically different result?
>>> >
>>> > How different? It's possible you are changing the numerical stability
>>> > of the expression. Or it's possible there is a bug in together.
>>> >
>>> > Aaron Meurer
>>> >
>>> >>
>>> >> I have a series of equations that I've been substituting in to each
>>> other
>>> >> building up a final symbolic result. My first pass at this gave a
>>> result
>>> >> that looks reasonable (though I haven't proven it correct) when I
>>> convert it
>>> >> to a numpy function after using
>>> >> Eq.subs(dictionary of values for symbols).simplify().n()
>>> >> and plot it.
>>> >>
>>> >> When I took the last symbolic equation to be substituted and use
>>> >> Eq.together() on it before substituting to get a simpler result, the
>>> >> Eq.subs().simplify() returns False. When I manipulate it differently
>>> to get
>>> >> it to successfully lambdify, the resulting plot is different (and
>>> appears
>>> >> wrong).
>>> >>
>>> >> Thanks--
>>> >>
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